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Introduction to Health Physics: Fourth Edition - Ruang Baca FMIPA UB

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568 CHAPTER 10<br />

energy equal <strong>to</strong> the maximum energy of the beta particles that gave rise <strong>to</strong> the Xrays.<br />

However, this quantum energy is, in fact, the upper limit of the bremsstrahlung<br />

energy, and all the bremsstrahlung is much lower in energy than this upper limit.<br />

The thickness calculated above, therefore, implicitly accounts for buildup through<br />

the use of an attenuation coefficient for the upper-energy limit of the X-rays rather<br />

than for their average energy.<br />

Neutrons<br />

Shielding against neutrons is based on slowing down fast neutrons and absorbing<br />

thermal neutrons. In Chapter 5, it was seen that attenuation and absorption of<br />

neutrons is a complex series of events. Despite the complexity, however, the required<br />

shielding around a neutron source can be estimated by the use of removal cross<br />

sections. (For neutron energies up <strong>to</strong> 30 MeV, the removal cross section is about<br />

three-quarters of the <strong>to</strong>tal cross section.) In designing shielding against neutrons, it<br />

must be borne in mind that absorption of neutrons can lead <strong>to</strong> induced radioactivity<br />

and <strong>to</strong> the production of gamma radiation (capture gammas) when the neutron is<br />

absorbed in<strong>to</strong> a nucleus.<br />

W Example 10.13<br />

Design a shield for an 18.5 × 10 4 MBq (5 Ci) Pu–Be neutron source that emits<br />

5 × 10 6 neutrons/s, such that the dose rate at the outside surface of the shield will<br />

not exceed 0.02 mSv/h (20 μSv/h or 2 mrems/h). The mean energy of the neutrons<br />

produced in this source is 4 MeV.<br />

Solution<br />

Let us make the shield of water and compute the minimum radius for the case<br />

of a spherical shield. Since we know that the capture of a neutron by hydrogen<br />

produces a 2.26-MeV gamma ray, let us allow for the gamma-ray dose by designing<br />

the shield <strong>to</strong> give a maximum fast-neutron dose rate of 0.01 mSv/h (1 mrem/h),<br />

which corresponds <strong>to</strong> a fast flux of 3.7 neutrons/cm2 /s (Table 9-5). The <strong>to</strong>tal cross<br />

section for 4-MeV neutrons for hydrogen and oxygen are 1.9 and 1.7 b, respectively.<br />

Since water contains 6.7 × 1022 hydrogen a<strong>to</strong>ms and 3.35 × 1022 oxygen a<strong>to</strong>ms per<br />

cm3 , the linear absorption coefficient, , of water is<br />

= 1.9 × 10 −24 cm2<br />

a<strong>to</strong>m<br />

= 0.187 cm −1 ,<br />

a<strong>to</strong>ms<br />

cm2<br />

a<strong>to</strong>ms<br />

× 6.7 × 1022 + 1.7 × 10−24 × 3.5 × 1022<br />

cm3 a<strong>to</strong>m cm3

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